问题
解答题
计算: (1)(
(2)
(3)
(4)若a是
|
答案
(1)原式=(2
-6
)-(2 2
+2 4
)6
=2
-6
-2 2
-2 4 6
=
-6
;3 2 4
(2)原式=
-22(2+
)3 4-3
+3 1
+13
=4+2
-23
+3
-13 2
=4+
-3 2 1 2
=3
+1 2
;3 2
(3)原式=
-b(
-b)ab ab-b2
+a(a+
)ab a2-ab a+b a-b
=
-
-bab a-b
+a+ ab a-b a+b a-b
=0;
(4)∵a是
的小数部分,2
∴a=
-1,2
∵b是a的倒数,
∴ab=1,b=
=1
-12
+1,2
+a b
-4=b a
=a2+b2-4ab ab
=(a-b)2-2ab ab
=2.(
-1-2
-1)2-22 1